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3 years ago

10

Using Models to Answer Questions About Systems

2 answers:

oksian1 [2.3K]3 years ago

8 0

^^ what he or she said

viva [34]3 years ago

5 0

1A: The legs can be a adjusted, as well as the sand can be swapped out. It’s a very good design for running multiple tests.
1B: He could add books or something under the front or back legs in order to increase/decrease the incline, therefore imitating the hypothesis.
1C: He can change out the sand grains to finer ones, or coarser ones, and record his results of each test.
2: If he sets the model at a steep incline and tests it with coarse sand and fine sand, seeing which one makes a narrower, deeper hole.

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How many liters of ammonia gas can be formed from 54.1 liters of hydrogen

Aleks04 [339]

<span>The reaction is N2 + 3H2 -> 2NH3
So the amount of NH3 formed is 2/3 of the amount of H2 = 2/3 * 13.7 = 9.13 Liters.</span><span>The answer is 9.13

</span>

5 0

3 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

<u>Answer:</u> The value of $K_{eq}$ is $4.84\times 10^{-5}$

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

<u>Initial:</u> 0.0280 0.0120

<u>At eqllm:</u> 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Answer:Molecular Formula SO2

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3 years ago

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Answer:

Explanation:

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